Secular resonances between bodies on close orbits II: prograde and retrograde orbits for irregular satellites

Abstract

In extending the analysis of the four secular resonances between close orbits in Li and Christou (Celest Mech Dyn Astron 125:133–160, 2016) (Paper I), we generalise the semianalytical model so that it applies to both prograde and retrograde orbits with a one-to-one map between the resonances in the two regimes. We propose the general form of the critical angle to be a linear combination of apsidal and nodal differences between the two orbits $b_1 \Delta \varpi + b_2 \Delta \varOmega$, forming a collection of secular resonances in which the ones studied in Paper I are among the strongest. Test of the model in the orbital vicinity of massive satellites with physical and orbital parameters similar to those of the irregular satellites Himalia at Jupiter and Phoebe at Saturn shows that ${>}20$ and ${>}40\%$ of phase space is affected by these resonances, respectively. The survivability of the resonances is confirmed using numerical integration of the full Newtonian equations of motion. We observe that the lowest order resonances with $b_1+|b_2|\le 3$ persist, while even higher-order resonances, up to $b_1+|b_2|\ge 7$, survive. Depending on the mass, between 10 and 60% of the integrated test particles are captured in these secular resonances, in agreement with the phase space analysis in the semianalytical model.

Appendix: Relation of orbital elements in the original and flipped frame

We derive the relation of the orbital elements in the two reference systems (1). Figure 13 shows the illustration of the frames and the quantities used in the derivation. As described, each of the frames can be transformed to the other by rotating it along the x-axis for $\pi$. We base the derivation on vector transformation in the two frames.

Fig. 13

Two reference frames: left: the “original” fame in which $i_\odot =0$; right: the “flipped” frame where $i_\odot '=\pi$ and where we can use the coorbital theory. The planet is at the origin of both frames. The outer dotted parallelogram is the orbital plane of the Sun (with respect to the planet), and the big solid ellipse is the orbit of the Sun; the smaller inner inclined one is the orbit of a retrograde satellite; the arrows mark the direction of Solar and satellite’s orbital revolution. ${\mathbf {r}}$, ${\mathbf {v}}$, ${\mathbf {h}}$ and ${\hat{\varOmega }}$ are the vectors of position, velocity, angular momentum and the unit vector representing the direction of ascending node in the two frames

Suppose ${\mathbf {P}}$ is an arbitrary vector and it is $[P_x, P_y, P_z]^\intercal$ in the original frame. Then in the flipped frame, it becomes

$$\begin{aligned} {\mathbf {P}}'=\left[ \begin{array}{c} P'_x\\ P'_y\\ P'_z\end{array}\right] ={\mathbf {R}}^{x}_\pi {\mathbf {P}}=\left[ \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} -1&{} 0 \\ 0 &{} 0 &{} -1 \end{array} \right] \left[ \begin{array}{c} P_x\\ P_y\\ P_z \end{array}\right] =\left[ \begin{array}{c} P_x\\ -P_y\\ -P_z\end{array}\right] , \end{aligned}$$

(8)

where ${\mathbf {R}}^{x}_\pi$ is the rotational matrix, meaning rotating the vector along x-axis for $\pi$; note that ${{\mathbf {R}}^{x}_\pi }^{-1}={\mathbf {R}}^{x}_\pi$.

First, we consider the position vector ${\mathbf {r}}$, the velocity vector ${\mathbf {v}}$ and the angular momentum vector ${\mathbf {h}}$. The notations without prime correspond to the original frame and primed ones relate to the flipped frame. Since the transition to the flipped frame is only a rotation, the length of the vectors remains conserved. Thus, Eqs. (2.134) and (2.135) of Murray and Dermott (1999) directly tell $a'=a$ and $e'=e$. According to the definition, the inclination $i'=\arccos (h_z'/\sqrt{h_x'^2+h_y'^2+h_z'^2}) =\arccos (-h_z/\sqrt{h_x^2+h_y^2+h_z^2})=\pi -i$.

The vector representing the direction of ascending node ${\hat{\varOmega }}$ is frame specific, and we show the calculation of it. Since ${\hat{\varOmega }}$ is in the orbital plane as well as in the $x-y$ plane, we have ${\hat{\varOmega }} \perp {\mathbf {h}}$ and ${\hat{\varOmega }} \perp {\hat{z}}$ (${\hat{z}}$ is the unit vertical vector). From the definition of ascending node, we know ${\hat{\varOmega }} ={\hat{z}} \times {\mathbf {h}}=[- h_y, h_x,0]^\intercal$. Thus, ${\hat{\varOmega }}' ={\hat{z}}' \times {\mathbf {h}}'=(- h_y', h_x',0)^\intercal =[ h_y, h_x,0]^\intercal$. Since $\varOmega$ is measured from x-axis, $\sin \varOmega =h_x/\sqrt{h_x^2+h_y^2}$ and $\cos \varOmega =-h_y/\sqrt{h_x^2+h_y^2}$. Thus,

$$\begin{aligned} \begin{aligned} \sin \varOmega '= & {} h_x'\big /\sqrt{h_x'^2+h_y'^2}= & {} h_x\big /\sqrt{h_x^2+h_y^2}= & {} \sin \varOmega \,\,\\ \cos \varOmega '= & {} -h_y'\big /\sqrt{h_x'^2+h_y'^2}= & {} h_y\big /\sqrt{h_x^2+h_y^2}= & {} -\cos \varOmega . \end{aligned} \end{aligned}$$

(9)

Hence, $\varOmega '=\pi -\varOmega$. Following Eq. (2.139) of Murray and Dermott (1999), f is a function of a, e, h (length of ${\mathbf {h}}$), r (length of ${\mathbf {r}}$) and $\dot{r}$, all remaining the same in the flipped frame, implying $f'=f$. Combining Eq. (2.138) with Eq. (8), we have $\sin (\omega '+f')=-\sin (\omega +f)$ and $\cos (\omega '+f')=-\cos (\omega +f)$. Hence, $\omega '+f'=\omega +f+\pi$ and $\omega '=\omega +\pi$.

Now we have constructed the relation between the orbital elements in the two frames (1).